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So suppose we have $X=\{(x,y)\in \mathbb{C} : y^2=x^2+x^3\}$ and i am asked to see that this is not a Riemann surface. Well we know that it is not going to be a smoth affine plane curve because at the point $(0,0)$ both the derivatives are $0$ but this is not enough, how can one ensure that there is possibly no way we cant find complex chart for this? I have tried supposing we can find a chart near $(0,0)$, so that we have $\phi^{-1}(z)=(\phi_1(z),\phi_2(z))$ near that point such that $\phi_2^2(z)=\phi_1^3(z)+\phi_1^2(z)$ but i cant seem to find a contradiction.Maybe seing something with the function $F$ that defines this locus set , since $\frac{\partial F}{\partial y} = 2\phi_2(z)\phi_2'(z)=0$ and $\frac{\partial F}{\partial x}=3\phi_1(z)^2\phi_1'(z)+2\phi_1(z)\phi_1'(z)=0$ so we get additional information that $3\phi_1(z)^2+2\phi_1(z)=0$, so that $\phi_1(z)=-\frac{2}{3}$ and this is a contadiction.Would this for example work ?

Thanks in advance.

Someone
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    Is $y^2=x^2$ a smooth curve in a neighborhood of the origin? Well, locally, this will be the union of two complex disks meeting at a point (just as you would have with $xy=0$). Is that homeomorphic to a single complex disk? – Ted Shifrin Apr 29 '20 at 21:58
  • No it will not , so yeah i guess that says so there cant be a complex chart because then it would be a homeomorphism are we would get a contradiction, altouhgt how do we see that locally that is the union of two complex disks meeting at a point? – Someone Apr 29 '20 at 22:02
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    Take the unit ball in $\Bbb C^2$ and intersect with $xy=0$. What do you get? Now do the same thing with $x^2=y^2$. – Ted Shifrin Apr 29 '20 at 22:11
  • Trying to follow past arguments … For the first intersection we get the union of ${(z,w)\in\mathbb{C}^2\vert z=0 \land |w|=1}$ and ${(z,w)\in\mathbb{C}^2\vert w=0 \land |z|=1}$. This union is homeomorphic to the union of the boundaries of two unit circles in $\mathbb{C}$ intersecting at one point. Correct? For the second intersection we get the union of ${(z,w)\in\mathbb{C}^2\vert |z|=2^{-1/2} \land z=w}$ and ${(z,w)\in\mathbb{C}^2\vert |z|=2^{-1/2}\land z=-w}$ — again homeomorphic to the union of the boundaries of two unit circles in $\mathbb{C}$ intersecting at one point, right? – Peter Jan 19 '23 at 10:09

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There are two pieces of smooth curve (= two Riemann surfaces) passing through $(0,0)$

$$y^2=x^2(1+x), \qquad y = \pm x \sqrt{1+x}$$

There is no "chart" from a neighborhood of $0$ to two copies of a neighborhood of $0$.

The function $$y+x\sqrt{1+x}$$ vanishes on one but not on the other which means that the ring of functions from the curve to $\Bbb{C}$ meromorphic near $(0,0)$ is not an integral domain (it is the product of two fields, two copies of the field of meromorphic functions near $0\in \Bbb{C}$).

reuns
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  • I dont think i get your point, could you elaborate a bit more pls? – Someone Apr 29 '20 at 21:54
  • Hm ok that makes sense but wouldnt take argument work with say $y^2=x+1$, we know that this is a smooth affine plane curve both we can have $y=\sqrt{x+1}$ and $y=-\sqrt{x+1}$, so that near the point $(-1,0)$ we would have the same problem? – Someone Apr 30 '20 at 07:36